1053 Path of Equal Weight (30 分)

Given a non-empty tree with root

Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed in the following figure: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: {10 5 2 7}, {10 4 10}, {10 3 3 6 2} and {10 3 3 6 2}, which correspond to the red edges in the figure.

Input Specification:

Each input file contains one test case. Each case starts with a line containing $0, the number of nodes in a tree,$

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 00.

Output Specification:

For each test case, print all the paths with weight S in non-increasing order. Each path occupies a line with printed weights from the root to the leaf in order. All the numbers must be separated by a space with no extra space at the end of the line.

Note: sequence ${ is said to be greater than sequence { if there exists 1 such that$

Sample Input:

20 9 24
10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2
00 4 01 02 03 04
02 1 05
04 2 06 07
03 3 11 12 13
06 1 09
07 2 08 10
16 1 15
13 3 14 16 17
17 2 18 19

Sample Output:

10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2


这题挺不错的，但是算是一个dfs遍历的模板题，没加啥东西。

 1 #include <bits/stdc++.h>
 2 #define ll long long int
 3 using namespace std;
 4 int n,m,k;
 5 struct Node
 6 {
 7     int to;
 8     ll val;
 9     friend bool operator < (const Node &a, const Node &b){
10         return a.val > b.val;
11     }
12 };
13 vector<Node> v[105];
14 int vis[105];
15 ll an[105];
16 int dp[105];
17 
18 
19 void output(int x){
20     stack<int> st;
21     while(x != 0){
22         st.push(an[x]);
23         x = dp[x];
24     }
25     st.push(an[0]);
26     while(!st.empty()){
27         cout << st.top();
28         if(st.size() == 1){
29             cout << endl;
30         }else{
31             cout << " ";
32         }
33         st.pop();
34     }
35 }
36 
37 void dfs(int x, ll sum){
38     vis[x] = 1;
39     if(sum == k){
40         if(v[x].size() == 0)
41             output(x);
42         return ;
43     }else if(sum > k){
44         return ;
45     }
46     for(int i = 0; i < v[x].size(); i++){
47         if(vis[v[x][i].to])
48             continue;
49         dp[v[x][i].to] = x;
50         dfs(v[x][i].to, sum+v[x][i].val);
51     }
52 }
53 
54 
55 int main(){
56     scanf("%d%d%d", &n, &m, &k);
57     for(int i = 0; i < n; i++){
58         cin >> an[i];
59     }
60     int x,y,z;
61     for(int i = 0; i < m; i++){
62         scanf("%d%d", &x,&y);
63         for(int j = 0; j < y; j++){
64             scanf("%d", &z);
65             v[x].push_back({z,an[z]});
66         }
67     }
68     for(int i = 0; i < n; i++){
69         sort(v[i].begin(), v[i].end());
70     }
71     dfs(0, an[0]);
72     return 0;
73 }